Indexing Results of a For Loop When the Input Comes from a Function

Question

Jeremy 2018-3-10

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https://ww2.mathworks.cn/matlabcentral/answers/387542-indexing-results-of-a-for-loop-when-the-input-comes-from-a-function

回答： Jeremy 2018-3-11

I just started using functions as loop inputs and never had to index the results before. My goal is to store the results of rpf_cba and vpf_cba into a matrix. Here's what I'm working with:

for i=0:h:11940
    [r_step_cba, v_step_cba] = RK4_single_step(mu, rpf_cba, vpf_cba, h);
    i = i+h;
    rpf_cba = r_step_cba;
    vpf_cba = v_step_cba;
    semi_cba = -(norm(rpf_cba)*mu)/((norm(rpf_cba)*norm(vpf_cba)^2)-(2*mu)).*semi_store
    i = i+1;
end

I'm quite used to indexing loop results and tried all of the usual methods (preallocation, index variable, etc.) but after spending a couple of days on this, I cannot figure out what's going on. The most common error I get is this:

In an assignment A(I) = B, the number of elements in B and I must be the same. Error in Exam_2_Question_4 (line 60) rpf_cba(i) = r_step_cba;

which makes me think the loop and function aren't looking in the same place for results to actually index. Alternatively, if I take the semi_ones variable and multiply it by r_step_cba, I get a 'Matrix dimensions must agree' error. If anyone has some experience with this, I'd be grateful for some advice. Also, let me know if more code is needed; the input function is pretty big so I didn't want to clutter things from the start. Thank you very much!

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Answer 1

Abraham Boayue 2018-3-11

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[r_step_cba, v_step_cba] = RK4_single_step(mu, rpf_cba, vpf_cba, h)
               for i = 0:h:11940 
                    i = i+h;
                    rpf_cba = r_step_cba;
                    vpf_cba = v_step_cba;
                   semi_cba = (norm(rpf_cba)*mu)/((norm(rpf_cba)*norm(vpf_cba)^2)-(2*mu))*semi_ones; 
                    i = i+1; 
                end

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Abraham Boayue 2018-3-11

Try to separate your comments from your code. It makes it difficult to understand the problem that you are having.

Jeremy 2018-3-11

Thanks for the heads up. I realized the "Code" button would do the trick.

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Answer 2

Jeremy 2018-3-11

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Nevermind. Just figured it out. In case this helps anyone, here's what I did...pretty simple, now that I think about it:

semi_store = [];
rpf_store = [];
vpf_store = [];
semi_ones = ones(1,200);
i = 0;
for t=0:step:11940
    [ r_step, v_step ] = RK4_single_step_cba_plus_drag(mu, rpf, vpf,...
        step);
    t = t+step;
    rpf = r_step;
    vpf = v_step;
    semi_drag = -(norm(rpf)*mu)/((norm(rpf)*norm(vpf)^2)-(2*mu));
    rpf_store = [rpf_store rpf]
    vpf_store = [vpf_store vpf];
    semi_store = [semi_store semi_drag];
    i = i+1;
end

I realized it is possible to pass an empty matrix into a for loop and use the results from the function to populate it. Now I'm sure for the savvy programmer, this was a "well, no duh" solution. For the rest of us, though, there you go.